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A three-point boundary value problem with an integral two-space-variables condition for parabolic equations
Computers and Mathematics with Applications 53 (2007) 940–947 www.elsevier.com/locate/camwa
A three-point boundary value problem with an integral two-space-variables condition for parabolic equations A.L. Marhoune Laboratoire Equations Differentielle, D´epartement de Mathematiques, Facult´e des Sciences, Universit´e Mentouri Constantine, 25000 Constantine, Algeria Received 25 July 2005; received in revised form 28 December 2005; accepted 10 April 2006
Abstract In this paper, we study a three-point boundary value problem with an integral two-space-variables condition for a class of parabolic equations. The existence and uniqueness of the solution in the functional weighted Sobolev space are proved. The proof is based on two-sided a priori estimates and on the density of the range of the operator generated by the problem considered. c 2007 Elsevier Ltd. All rights reserved.
Keywords: Integral boundary two-space-variables condition; Energy inequalities; Three-point boundary conditions; Equation of mixed type; Weighted Sobolev spaces
1. Introduction In the rectangle Ω = (0, 1) × (0, T ), we consider the equation Lu =
∂ 2u ∂u − a (t) 2 = f (x, t) ∂t ∂x
(1)
where the function a (t) and its derivative are bounded on the interval [0, T ]: 0 < a0 < a (t) ≤ a1 , da (t) 0 < a2 ≤ ≤ a3 . dt To Eq. (1) we add the initial condition lu = u(x, 0) = ϕ(x),
x ∈ (0, 1),
(2)
the boundary condition u(0, t) = u(1, t)
t ∈ (0, T ),
E-mail address: ahmed [email protected]. c 2007 Elsevier Ltd. All rights reserved. 0898-1221/$ - see front matter doi:10.1016/j.camwa.2006.04.031
(3)
A.L. Marhoune / Computers and Mathematics with Applications 53 (2007) 940–947
and the integral condition Z 1 Z α u(ξ, t)dξ = 0 α > 0, β > 0, α + β = 1 u(ξ, t)dξ + β
0
t ∈ (0, T ).
941
(4)
Here, we assumed that the known function ϕ satisfies the conditions given in (3) and (4), i.e., Z 1 Z α ϕ(0) = ϕ(1), ϕ(x)dx = 0. ϕ(x)dx + β
0
When considering the classical solution of the problem (3) and (4), along with the conditions (4) there should be fulfilled the conditions a(0) ϕ 00 (1) − ϕ 00 (0) = f (1, 0) − f (0, 0) ) Z (Z Z 1 Z 1 α α 00 00 f (x, 0)dx = 0. f (x, 0)dx + ϕ (x)dx + ϕ (x)dx + a(0) β
0
0
β
The importance of boundary value problems with integral boundary conditions has been pointed out by Samarskii [1]. We remark that integral boundary conditions for evolution problems have various applications in chemical engineering, thermoelasticity, underground water flow and population dynamics; see for example [2–5]. Boundary value problems for parabolic equations with an integral boundary condition are investigated by Batten Jr. [6], Bouziani and Benouar [7], Cannon [8,9], Cannon, Perez Esteva and Van Der Hoeck [10], Ionkin [11], Kamynin [12], Shi and Shillor [13], Shi [4], Marhoune and Bouzit [14], Denche and Marhoune [15–17], Yurchuk [18], and many references therein. The problem with an integral one-space-variable condition is studied in Kartynnik [19], Denche and Marhoune [20]. The present paper is devoted to the study of a problem with a boundary integral two-space-variables condition for a second-order partial differential equation. We associate with problem (1)–(4) the operator L = (L, l), defined from E into F, where E is the Banach space of functions u ∈ L 2 (Ω ), satisfying (3) and (4), with the finite norm " 2 2 # Z 1 Z 2 ∂ u θ (x) ∂u 2 ∂u θ (x) 2 dx kuk E = ∂t + ∂ x 2 dxdt + sup 2 ∂x Ω 2 0≤t≤T 0 Z α Z 1 + sup |u|2 dx + sup |u|2 dx (5) 0≤t≤T
0
0≤t≤T
β
and F is the Hilbert space of vector-valued functions F = ( f, ϕ) obtained by completion of the space L 2 (Ω ) × W22 (0, 1) with respect to the norm Z 1 Z 1 Z α Z θ (x) dϕ 2 θ (x) 2 |ϕ|2 dx | f |2 dxdt + |ϕ| dx + dx + (6) kFk2F = k( f, ϕ)k2F = dx 2 2 0 0 β Ω where 2 x θ (x) = (1 − β)2 (1 − x)2
0
Using the energy inequalities method proposed in [18], we establish two-sided a priori estimates. Then, we prove that the operator L is a linear homeomorphism between the spaces E and F. 2. Two-sided a priori estimates Theorem 1. For any function u ∈ E we have the a priori estimate kLuk F ≤ ckuk E , where the constant c is independent of u.
(7)
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A.L. Marhoune / Computers and Mathematics with Applications 53 (2007) 940–947
Proof. Using Eq. (1) and initial conditions (2) we obtain " 2 2 # Z Z θ (x) θ (x) ∂u 2 2 2 ∂ u |Lu| dxdt ≤ 2 + a1 2 dxdt, ∂t ∂x Ω 2 Ω 2 Z 1 Z 1 θ (x) dϕ 2 θ (x) ∂u 2 dx ≤ sup dx, 2 dx 2 ∂x 0 0≤t≤T 0 Z α Z α 2 |u|2 dx, |ϕ| dx ≤ sup 0
(8)
(9) (10)
0
0≤t≤T
and Z β
1
|ϕ|2 dx ≤ sup 0≤t≤T
Z β
1
|u|2 dx.
(11)
Combining the inequalities (8)–(11), we obtain (7) for u ∈ E.
Theorem 2. For any function u ∈ E, we have the a priori estimate kuk E ≤ kkLuk F ,
(12)
with the constant exp(cT ) max 33, a20 , k= 2 7 a0 a0 min 16 , 2, 2 and c is such that ca0 ≥ a3 .
(13)
Before proving this theorem, we first give the following two lemmas. Lemma 3 ([21]). For u ∈ E, we have Z α 2 Z Z ∂u 1 α α ∂u(ξ, t) 2 dx ≤ x 2 dx. dξ 4 0 ∂t ∂t x 0
(14)
Lemma 4 ([21]). For u ∈ E, we have 2 Z Z Z 1 1 1 x ∂u(ξ, t) 2 2 ∂u dξ dx ≤ − x) (1 ∂t dx. 4 β β ∂t β Proof of Theorem 2. Define Z α ∂u(ξ, t) 2 ∂u x + 2x dξ ∂t ∂t x ∂u Mu = (1 − β)2 ∂t Z x ∂u(ξ, t) 2 ∂u + 2 (1 − x) dξ (1 − x) ∂t ∂t β We consider for u ∈ E the quadratic formula Z τZ 1 Re exp (−ct) Lu Mudxdt, 0
0
(15)
0
(16)
A.L. Marhoune / Computers and Mathematics with Applications 53 (2007) 940–947
943
with the constant c satisfying (13), obtained by multiplying Eq. (1) by exp(−ct)Mu, by integrating over Ω τ , where Ω τ = (0, 1)×(0, τ ), with 0 ≤ τ ≤ T , and by taking the real part. Integrating by parts in (16) with the use of boundary conditions (3) and (4), we obtain 2 Z τZ 1 Z τZ 1 ∂u exp (−ct) Lu Mudxdt = θ (x) exp (−ct) dxdt Re ∂t 0 0 0 0 Z α Z τZ α 1 ∂u(ξ, t)dξ 2 + exp (−ct) dxdt 2 0 0 ∂t x Z x Z Z ∂u(ξ, t)dξ 2 1 τ 1 exp (−ct) + dxdt 2 0 β ∂t β Z τZ 1 ∂u ∂ 2 u dxdt exp (−ct) θ (x)a + Re ∂ x ∂ x∂t 0 0 Z Z τ α ∂u + 2Re exp (−ct) ua dxdt ∂t 0 0 Z τZ 1 ∂u exp (−ct) au dxdt. + 2Re (17) ∂t 0 β On the other hand, by using the elementary inequalities we get 2 Z τZ 1 Z τZ 1 ∂u Re exp (−ct) Lu Mudxdt ≥ θ (x) exp (−ct) dxdt ∂t 0 0 0 0 Z τZ 1 ∂u ∂ 2 u dxdt + Re exp (−ct) θ (x)a ∂ x ∂ x∂t 0 0 Z τZ α ∂u + 2Re exp (−ct) ua dxdt ∂t 0 0 Z τZ 1 ∂u + 2Re exp (−ct) au dxdt. ∂t 0 β
(18)
Again, integrating by parts the second, third and fourth terms of the right-hand side of the inequality (18) and taking into account the initial conditions (2) and (13) gives 2 Z τZ 1 Z ∂u 1 1 ∂u ∂ 2 u exp (−cτ ) θ (x)a(τ ) (x, τ ) dx Re dxdt ≥ exp (−ct) θ (x)a ∂ x ∂ x∂t 2 0 ∂x 0 0 Z dlu 2 1 1 dx − θ (x)a(0) (19) 2 0 dx Z τZ α Z Z ∂u 1 α 1 α Re exp (−ct) ua dxdt ≥ exp (−cτ ) a(τ ) |u(x, τ )|2 dx − a(0) |lu|2 dx (20) ∂t 2 2 0 0 0 0 Z τZ 1 Z Z ∂u 1 1 1 1 exp (−ct) au dxdt ≥ exp (−cτ ) a(τ ) |u(x, τ )|2 dx − a(0) |lu|2 dx. (21) Re ∂t 2 β 2 β 0 β Using (19)–(21) in (18) we get Z Z τZ ` Re exp (−ct) Lu Mudxdt + 0
a(0) |lu|2 dx +
0
0 τ
α
1
Z β
a(0) |lu|2 dx +
1 2
1
Z 0
dlu 2 dx θ (x)a(0) dx
2 2 Z ∂u ∂u 1 1 ≥ θ (x) exp (−ct) dxdt + exp (−cτ ) θ (x)a(τ ) (x, τ ) dx ∂t 2 0 ∂x 0 0 Z α Z 1 + exp (−cτ ) a(τ ) |u(x, τ )|2 dx + exp (−cτ ) a(τ ) |u(x, τ )|2 dx. Z
Z
0
1
β
(22)
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A.L. Marhoune / Computers and Mathematics with Applications 53 (2007) 940–947
By using the elementary inequalities on the first integral in the left-hand side of (22) and Lemmas 3 and 4 we obtain 2 2 Z Z Z ∂u ∂u 23 τ 1 1 1 θ (x) exp (−ct) dxdt + exp (−cτ ) θ (x)a(τ ) (x, τ ) dx 32 0 0 ∂t 2 0 ∂x Z α Z 1 + exp (−cτ ) a(τ ) |u(x, τ )|2 dx + exp (−cτ ) a(τ ) |u(x, τ )|2 dx β
0 τ
Z
1
Z
≤ 16 0 α
Z ×
θ (x) exp (−ct) |Lu|2 dxdt +
0
a(0) |`u|2 dx +
0
1
Z β
1 2
1
Z 0
d`u 2 dx θ (x)a(0) dx
a(0) |`u|2 dx.
Now, from Eq. (1) we have 2 2 Z Z Z Z ∂ u 1 τ 1 a0 τ 1 θ (x) exp (−ct) 2 dxdt ≤ exp (−ct) θ (x) |Lu|2 dxdt 2 0 0 2 0 0 ∂x 2 Z Z ∂u 1 τ 1 exp (−ct) θ (x) dxdt. + 2 0 0 ∂t Combining inequalities (23) and (24) we get Z Z Z α Z dlu 2 33 a1 1 a0 1 2 dx + a1 |lu| |lu|2 dx θ (x) |Lu|2 dxdt + θ (x) dx + 2 Ω 2 0 dx 2 0 2 β 2 2 Z τZ 1 Z ∂u ∂u 7 a0 1 ≥ exp(−cT ) θ (x) dxdt + θ (x) (x, τ ) dx 32 0 0 ∂t 2 0 ∂x Z α Z 1 a0 a0 |u (x, τ )|2 dx + |u (x, τ )|2 dx + 2 0 2 β ! 2 2 Z Z ∂ u a02 τ 1 θ (x) 2 dxdt . + 4 0 0 ∂x
(23)
(24)
(25)
As the left-hand side of (25) is independent of τ , by replacing the right-hand side by its upper bound with respect to τ in the interval [0, T ], we obtain the desired inequality. 3. Solvability of the problem From estimates (7) and (12) it follows that the operator L : E −→ F is continuous and its range is closed in F. Therefore, the inverse operator L −1 exists and is continuous from the closed subspace R (L) onto E, which means that L is a homeomorphism from E onto R (L). To obtain the uniqueness of the solution, it remains to show that R (L) = F. The proof is based on the following lemma. Lemma 5. Let D0 (L) = {u ∈ E : lu = 0}. If for u ∈ D0 (L) and some ω ∈ L 2 (Ω ), we have Z φ(x)Lu$ dxdt = 0, Ω
where x φ(x) = (1 − β) (1 − x) then ω = 0.
0
(26)
A.L. Marhoune / Computers and Mathematics with Applications 53 (2007) 940–947
Proof. From (26) we have Z Z ∂u ∂ 2u φ(x) $ dxdt = φ(x)a(t) 2 $ dxdt. ∂t ∂x Ω Ω
945
(27)
Now, for given ω (x, t), we introduce the function Z α ω(ξ, t) ω − dξ 0 < x ≤ α ξ x α≤x ≤β v (x, t) = ω Z x ω(ξ, t) ω − dξ β ≤ x < 1. β 1−ξ Integrating by parts with respect to ξ , we obtain Z α v(ξ, t)dξ 0
which implies that Z α Z v(ξ, t)dξ = x
1 β
v(ξ, t)dξ = 0.
(28)
Then, from equality (27) we obtain Z Z ∂u N vdxdt = − A (t) uvdxdt, Ω Ω ∂t
(29)
where N v = φ(x)v,
and
A (t) u = −
∂ ∂x
φ(x)a (t)
∂u ∂x
.
If we introduce the smoothing operators with respect to t [18,22–24], Jξ−1 = I + ξ ∂t∂ operators provide the solutions of the respective problems: dgξ (t) + gξ (t) = g (t) dt gξ (t) |t=0 = 0
ξ
−1
∗ and Jξ−1 , then these
(30)
and dgξ∗ (t)
+ gξ∗ (t) = g (t) dt gξ (t) |t=T = 0
−ξ
(31)
∗ and also have the following properties: for any g ∈ L 2 (0, T ), the functions gξ = Jξ−1 g and gξ∗ = Jξ−1 g are 2 RT in W21 (0, T ) such that gξ |t=0 = 0 and gξ∗ |t=T = 0. Moreover, Jξ−1 commutes with ∂t∂ , so 0 gξ − g dt −→ 0 and 2 R T ∗ − g g dt −→ 0 for ξ −→ 0. ξ 0 Rt ξ a2 Putting u = 0 exp(cτ )vξ∗ (x, τ ) dτ in (29), where the constant c satisfies ca0 − a3 − 2a30 ≥ 0, and using (31), we obtain Z Z Z ∂vξ∗ ∂u dxdt. (32) − exp(ct)vξ∗ N vdxdt = + A (t) u exp(−ct) dxdt − ξ A (t) u ∂t ∂t Ω Ω Ω
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A.L. Marhoune / Computers and Mathematics with Applications 53 (2007) 940–947
Integrating by parts each term in the left-hand side of (32) and taking the real parts yields 2 Z Z 1 ∂u ∂u A (t) u exp(−ct) dxdt = a (t) exp(−ct)φ(x) (x, T ) dx 2Re ∂t ∂t Ω 0 Z da(t) ∂u 2 exp(−ct)φ(x) ca (t) − + ∂ x dxdt, dt Ω ! ! Z Z ∗ ∂vξ∗ da(t) ∂u ∂vξ A (t) u Re −ξ dxdt = Re ξ φ(x) dxdt ∂t ∂x ∂x Ω Ω dt Z ∂v ∗ 2 ξ a (t) exp(−ct)φ(x) + dxdt. ∂x Ω
(33)
(34)
Using ε-inequalities we obtain Z Re −ξ Ω
A (t) u
∂vξ∗ ∂t
! dxdt
≥
Z da(t) 2 ∂u 2 −ξ dxdt. exp(−ct)φ(x) 2a0 Ω dt ∂ x
(35)
Combining (33) and (35) we get Z −Re Ω
exp(ct)vξ∗ N vdxdt
ξ a2 ≥ exp(−ct)φ(x) ca0 − a3 − 3 2a0 Ω Z
! ∂u 2 dxdt ≥ 0. ∂x
(36)
Now, using (36), we have Z Re exp(ct)vξ∗ N vdxdt ≤ 0. Ω
Then, for ξ −→ 0 we obtain Z Z Re exp(ct)v N vdxdt = exp(ct)φ(x) |v|2 dxdt ≤ 0. Ω
Ω
We conclude that v = 0; hence, ω = 0, which ends the proof of the lemma.
Theorem 6. The range R(L) of L coincides with F. Proof. Since F is a Hilbert space, we have R(L) = F if and only if the relation Z 1 Z 1 Z α Z θ (x) θ (x) dlu dϕ Lu f dxdt + dx + luϕdx + luϕdx = 0 2 2 dx dx Ω 0 0 β
(37)
for arbitrary u ∈ E and ( f, ϕ, ) ∈ F, implies that f = 0 and ϕ = 0. Putting u ∈ D0 (L) in (37), we conclude from the Lemma 3 that ψ f = 0, then f = 0, where 0
θ(x) dϕ 2 2 | dx |
+
Rα 0
|ϕ|2 dx
A.L. Marhoune / Computers and Mathematics with Applications 53 (2007) 940–947
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References [1] A.A. Samarskii, Some problems in differential equations theory, Differ. Uravn. 16 (11) (1980) 1221–1228. [2] Y.S. Choi, K.Y. Chan, A parabolic equation with nonlocal boundary conditions arising from electrochemistry, Nonlinear Anal. 18 (1992) 317–331. [3] R.E. Ewing, T. Lin, A class of parameter estimation techniques for fluid flow in porous media, Adv. Water Ressources 14 (1991) 89–97. [4] P. Shi, Weak solution to evolution problem with a nonlocal constraint, Siam J. Anal. 24 (1993) 46–58. [5] A.V. Kartynnik, Three-point boundary-value problem with an integral space-variable condition for a second-order parabolic equation, Differ. Equ. 26 (1990) 1160–1166. [6] G.W. Batten Jr., Second-order correct boundary conditions for the numerical solution of the mixed boundary problem for parabolic equations, Math. Comput. 17 (1963) 405–413. [7] A. Bouziani, N.E. Benouar, Mixed problem with integral conditions for a third order parabolic equation, Kobe J. Math. 15 (1998) 47–58. [8] J.R. Cannon, The solution of the heat equation subject to the specification of energy, Quart. Appl. Math. 21 (1963) 155–160. [9] J.R. Cannon, The one-dimensional heat equation, in: Encyclopedia of Mathematics and its Applications, vol. 23, Addison-Wesley, Mento Park, CA, 1984. [10] J.R. Cannon, S. Perez Esteva, J. Van Der Hoek, A Galerkin procedure for the diffusion equation subject to the specification of mass, Siam J. Numer. Anal. 24 (1987) 499–515. [11] N.I. Ionkin, Solution of a boundary-value problem in heat condition with a nonclassical boundary condition, Differ. Uravn. 13 (1977) 294–304. [12] N.I. Kamynin, A boundary value problem in the theory of the heat condition with non classical boundary condition, U.S.S.R. Comput. Math. Math. Phys. 4 (1964) 33–59. [13] P. Shi, M. Shillor, Design of Contact Patterns in One Dimensional Thermoelasticity, in: Theoretical Aspects of Industrial Design, Society for Industrial and Applied Mathematics, Philadelphia, PA, 1992. [14] A.L. Marhoune, M. Bouzit, High order differential equations with integral boundary condition, Far East J. Math. Sci. 18 (3) (2005) 341–450. [15] M. Denche, A.L. Marhoune, Mixed problem with integral boundary condition for a high order mixed type partial differential equation, J. Appl. Math. Stochastic Anal. 16 (1) (2003) 69–79. [16] M. Denche, A.L. Marhoune, Mixed problem with nonlocal boundary conditions for a third-order partial differential equation of mixed type, Int. J. Math. Math. Sci. 26 (7) (2001) 417–426. [17] M. Denche, A.L. Marhoune, High-order mixed type partial differential equations with integral boundary conditions, Electron. J. Differ. Equ. 2000 (60) (2000) 1–10. [18] N.I. Yurchuk, Mixed problem with an integral condition for certain parabolic equations, Differ. Equ. 22 (1986) 1457–1463. [19] A.V. Kartynnik, Three-point boundary-value problem with an integral space-variable condition for a second-order parabolic equation, Differ. Equ. 26 (9) (1990) 1160–1166. [20] M. Denche, A.L. Marhoune, A three-point boundary value problem with an Integral condition for parabolic equations with the Bessel operator, Appl. Math. Lett. 13 (2000) 85–89. [21] G.H. Hardy, J.E. Littlewood, G. Polya, Inequalities, Cambridge Press, 1934. Zbl 0010.10703 | JFM 60.0169.01. [22] J. Pr¨uss, G. Simonett, Maximal regularity for evolution equations in weighted Lp-spaces, Fachbereich Mathematik und Informatik, MartinLuter-Universit¨at Halle-Wittenberg, July 2002. [23] M. Hieber, J. Pr¨uss, Heat Kernels and Maximal Lp–Lq estimates for parabolic evolution equations, Comm. Partial Differ. Equ. 22 (1997) 164761669. [24] P. Cannarasa, V. Vespri, On maximal Lp-regularity for abstract Cauchy problem, Boll. Unione Mat. Italiana (1986) 165–175.
A three-point boundary value problem with an integral two-space-variables condition for parabolic equations A.L. Marhoune Laboratoire Equations Differentielle, D´epartement de Mathematiques, Facult´e des Sciences, Universit´e Mentouri Constantine, 25000 Constantine, Algeria Received 25 July 2005; received in revised form 28 December 2005; accepted 10 April 2006
Abstract In this paper, we study a three-point boundary value problem with an integral two-space-variables condition for a class of parabolic equations. The existence and uniqueness of the solution in the functional weighted Sobolev space are proved. The proof is based on two-sided a priori estimates and on the density of the range of the operator generated by the problem considered. c 2007 Elsevier Ltd. All rights reserved.
Keywords: Integral boundary two-space-variables condition; Energy inequalities; Three-point boundary conditions; Equation of mixed type; Weighted Sobolev spaces
1. Introduction In the rectangle Ω = (0, 1) × (0, T ), we consider the equation Lu =
∂ 2u ∂u − a (t) 2 = f (x, t) ∂t ∂x
(1)
where the function a (t) and its derivative are bounded on the interval [0, T ]: 0 < a0 < a (t) ≤ a1 , da (t) 0 < a2 ≤ ≤ a3 . dt To Eq. (1) we add the initial condition lu = u(x, 0) = ϕ(x),
x ∈ (0, 1),
(2)
the boundary condition u(0, t) = u(1, t)
t ∈ (0, T ),
E-mail address: ahmed [email protected]. c 2007 Elsevier Ltd. All rights reserved. 0898-1221/$ - see front matter doi:10.1016/j.camwa.2006.04.031
(3)
A.L. Marhoune / Computers and Mathematics with Applications 53 (2007) 940–947
and the integral condition Z 1 Z α u(ξ, t)dξ = 0 α > 0, β > 0, α + β = 1 u(ξ, t)dξ + β
0
t ∈ (0, T ).
941
(4)
Here, we assumed that the known function ϕ satisfies the conditions given in (3) and (4), i.e., Z 1 Z α ϕ(0) = ϕ(1), ϕ(x)dx = 0. ϕ(x)dx + β
0
When considering the classical solution of the problem (3) and (4), along with the conditions (4) there should be fulfilled the conditions a(0) ϕ 00 (1) − ϕ 00 (0) = f (1, 0) − f (0, 0) ) Z (Z Z 1 Z 1 α α 00 00 f (x, 0)dx = 0. f (x, 0)dx + ϕ (x)dx + ϕ (x)dx + a(0) β
0
0
β
The importance of boundary value problems with integral boundary conditions has been pointed out by Samarskii [1]. We remark that integral boundary conditions for evolution problems have various applications in chemical engineering, thermoelasticity, underground water flow and population dynamics; see for example [2–5]. Boundary value problems for parabolic equations with an integral boundary condition are investigated by Batten Jr. [6], Bouziani and Benouar [7], Cannon [8,9], Cannon, Perez Esteva and Van Der Hoeck [10], Ionkin [11], Kamynin [12], Shi and Shillor [13], Shi [4], Marhoune and Bouzit [14], Denche and Marhoune [15–17], Yurchuk [18], and many references therein. The problem with an integral one-space-variable condition is studied in Kartynnik [19], Denche and Marhoune [20]. The present paper is devoted to the study of a problem with a boundary integral two-space-variables condition for a second-order partial differential equation. We associate with problem (1)–(4) the operator L = (L, l), defined from E into F, where E is the Banach space of functions u ∈ L 2 (Ω ), satisfying (3) and (4), with the finite norm " 2 2 # Z 1 Z 2 ∂ u θ (x) ∂u 2 ∂u θ (x) 2 dx kuk E = ∂t + ∂ x 2 dxdt + sup 2 ∂x Ω 2 0≤t≤T 0 Z α Z 1 + sup |u|2 dx + sup |u|2 dx (5) 0≤t≤T
0
0≤t≤T
β
and F is the Hilbert space of vector-valued functions F = ( f, ϕ) obtained by completion of the space L 2 (Ω ) × W22 (0, 1) with respect to the norm Z 1 Z 1 Z α Z θ (x) dϕ 2 θ (x) 2 |ϕ|2 dx | f |2 dxdt + |ϕ| dx + dx + (6) kFk2F = k( f, ϕ)k2F = dx 2 2 0 0 β Ω where 2 x θ (x) = (1 − β)2 (1 − x)2
0
Using the energy inequalities method proposed in [18], we establish two-sided a priori estimates. Then, we prove that the operator L is a linear homeomorphism between the spaces E and F. 2. Two-sided a priori estimates Theorem 1. For any function u ∈ E we have the a priori estimate kLuk F ≤ ckuk E , where the constant c is independent of u.
(7)
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A.L. Marhoune / Computers and Mathematics with Applications 53 (2007) 940–947
Proof. Using Eq. (1) and initial conditions (2) we obtain " 2 2 # Z Z θ (x) θ (x) ∂u 2 2 2 ∂ u |Lu| dxdt ≤ 2 + a1 2 dxdt, ∂t ∂x Ω 2 Ω 2 Z 1 Z 1 θ (x) dϕ 2 θ (x) ∂u 2 dx ≤ sup dx, 2 dx 2 ∂x 0 0≤t≤T 0 Z α Z α 2 |u|2 dx, |ϕ| dx ≤ sup 0
(8)
(9) (10)
0
0≤t≤T
and Z β
1
|ϕ|2 dx ≤ sup 0≤t≤T
Z β
1
|u|2 dx.
(11)
Combining the inequalities (8)–(11), we obtain (7) for u ∈ E.
Theorem 2. For any function u ∈ E, we have the a priori estimate kuk E ≤ kkLuk F ,
(12)
with the constant exp(cT ) max 33, a20 , k= 2 7 a0 a0 min 16 , 2, 2 and c is such that ca0 ≥ a3 .
(13)
Before proving this theorem, we first give the following two lemmas. Lemma 3 ([21]). For u ∈ E, we have Z α 2 Z Z ∂u 1 α α ∂u(ξ, t) 2 dx ≤ x 2 dx. dξ 4 0 ∂t ∂t x 0
(14)
Lemma 4 ([21]). For u ∈ E, we have 2 Z Z Z 1 1 1 x ∂u(ξ, t) 2 2 ∂u dξ dx ≤ − x) (1 ∂t dx. 4 β β ∂t β Proof of Theorem 2. Define Z α ∂u(ξ, t) 2 ∂u x + 2x dξ ∂t ∂t x ∂u Mu = (1 − β)2 ∂t Z x ∂u(ξ, t) 2 ∂u + 2 (1 − x) dξ (1 − x) ∂t ∂t β We consider for u ∈ E the quadratic formula Z τZ 1 Re exp (−ct) Lu Mudxdt, 0
0
(15)
0
(16)
A.L. Marhoune / Computers and Mathematics with Applications 53 (2007) 940–947
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with the constant c satisfying (13), obtained by multiplying Eq. (1) by exp(−ct)Mu, by integrating over Ω τ , where Ω τ = (0, 1)×(0, τ ), with 0 ≤ τ ≤ T , and by taking the real part. Integrating by parts in (16) with the use of boundary conditions (3) and (4), we obtain 2 Z τZ 1 Z τZ 1 ∂u exp (−ct) Lu Mudxdt = θ (x) exp (−ct) dxdt Re ∂t 0 0 0 0 Z α Z τZ α 1 ∂u(ξ, t)dξ 2 + exp (−ct) dxdt 2 0 0 ∂t x Z x Z Z ∂u(ξ, t)dξ 2 1 τ 1 exp (−ct) + dxdt 2 0 β ∂t β Z τZ 1 ∂u ∂ 2 u dxdt exp (−ct) θ (x)a + Re ∂ x ∂ x∂t 0 0 Z Z τ α ∂u + 2Re exp (−ct) ua dxdt ∂t 0 0 Z τZ 1 ∂u exp (−ct) au dxdt. + 2Re (17) ∂t 0 β On the other hand, by using the elementary inequalities we get 2 Z τZ 1 Z τZ 1 ∂u Re exp (−ct) Lu Mudxdt ≥ θ (x) exp (−ct) dxdt ∂t 0 0 0 0 Z τZ 1 ∂u ∂ 2 u dxdt + Re exp (−ct) θ (x)a ∂ x ∂ x∂t 0 0 Z τZ α ∂u + 2Re exp (−ct) ua dxdt ∂t 0 0 Z τZ 1 ∂u + 2Re exp (−ct) au dxdt. ∂t 0 β
(18)
Again, integrating by parts the second, third and fourth terms of the right-hand side of the inequality (18) and taking into account the initial conditions (2) and (13) gives 2 Z τZ 1 Z ∂u 1 1 ∂u ∂ 2 u exp (−cτ ) θ (x)a(τ ) (x, τ ) dx Re dxdt ≥ exp (−ct) θ (x)a ∂ x ∂ x∂t 2 0 ∂x 0 0 Z dlu 2 1 1 dx − θ (x)a(0) (19) 2 0 dx Z τZ α Z Z ∂u 1 α 1 α Re exp (−ct) ua dxdt ≥ exp (−cτ ) a(τ ) |u(x, τ )|2 dx − a(0) |lu|2 dx (20) ∂t 2 2 0 0 0 0 Z τZ 1 Z Z ∂u 1 1 1 1 exp (−ct) au dxdt ≥ exp (−cτ ) a(τ ) |u(x, τ )|2 dx − a(0) |lu|2 dx. (21) Re ∂t 2 β 2 β 0 β Using (19)–(21) in (18) we get Z Z τZ ` Re exp (−ct) Lu Mudxdt + 0
a(0) |lu|2 dx +
0
0 τ
α
1
Z β
a(0) |lu|2 dx +
1 2
1
Z 0
dlu 2 dx θ (x)a(0) dx
2 2 Z ∂u ∂u 1 1 ≥ θ (x) exp (−ct) dxdt + exp (−cτ ) θ (x)a(τ ) (x, τ ) dx ∂t 2 0 ∂x 0 0 Z α Z 1 + exp (−cτ ) a(τ ) |u(x, τ )|2 dx + exp (−cτ ) a(τ ) |u(x, τ )|2 dx. Z
Z
0
1
β
(22)
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A.L. Marhoune / Computers and Mathematics with Applications 53 (2007) 940–947
By using the elementary inequalities on the first integral in the left-hand side of (22) and Lemmas 3 and 4 we obtain 2 2 Z Z Z ∂u ∂u 23 τ 1 1 1 θ (x) exp (−ct) dxdt + exp (−cτ ) θ (x)a(τ ) (x, τ ) dx 32 0 0 ∂t 2 0 ∂x Z α Z 1 + exp (−cτ ) a(τ ) |u(x, τ )|2 dx + exp (−cτ ) a(τ ) |u(x, τ )|2 dx β
0 τ
Z
1
Z
≤ 16 0 α
Z ×
θ (x) exp (−ct) |Lu|2 dxdt +
0
a(0) |`u|2 dx +
0
1
Z β
1 2
1
Z 0
d`u 2 dx θ (x)a(0) dx
a(0) |`u|2 dx.
Now, from Eq. (1) we have 2 2 Z Z Z Z ∂ u 1 τ 1 a0 τ 1 θ (x) exp (−ct) 2 dxdt ≤ exp (−ct) θ (x) |Lu|2 dxdt 2 0 0 2 0 0 ∂x 2 Z Z ∂u 1 τ 1 exp (−ct) θ (x) dxdt. + 2 0 0 ∂t Combining inequalities (23) and (24) we get Z Z Z α Z dlu 2 33 a1 1 a0 1 2 dx + a1 |lu| |lu|2 dx θ (x) |Lu|2 dxdt + θ (x) dx + 2 Ω 2 0 dx 2 0 2 β 2 2 Z τZ 1 Z ∂u ∂u 7 a0 1 ≥ exp(−cT ) θ (x) dxdt + θ (x) (x, τ ) dx 32 0 0 ∂t 2 0 ∂x Z α Z 1 a0 a0 |u (x, τ )|2 dx + |u (x, τ )|2 dx + 2 0 2 β ! 2 2 Z Z ∂ u a02 τ 1 θ (x) 2 dxdt . + 4 0 0 ∂x
(23)
(24)
(25)
As the left-hand side of (25) is independent of τ , by replacing the right-hand side by its upper bound with respect to τ in the interval [0, T ], we obtain the desired inequality. 3. Solvability of the problem From estimates (7) and (12) it follows that the operator L : E −→ F is continuous and its range is closed in F. Therefore, the inverse operator L −1 exists and is continuous from the closed subspace R (L) onto E, which means that L is a homeomorphism from E onto R (L). To obtain the uniqueness of the solution, it remains to show that R (L) = F. The proof is based on the following lemma. Lemma 5. Let D0 (L) = {u ∈ E : lu = 0}. If for u ∈ D0 (L) and some ω ∈ L 2 (Ω ), we have Z φ(x)Lu$ dxdt = 0, Ω
where x φ(x) = (1 − β) (1 − x) then ω = 0.
0
(26)
A.L. Marhoune / Computers and Mathematics with Applications 53 (2007) 940–947
Proof. From (26) we have Z Z ∂u ∂ 2u φ(x) $ dxdt = φ(x)a(t) 2 $ dxdt. ∂t ∂x Ω Ω
945
(27)
Now, for given ω (x, t), we introduce the function Z α ω(ξ, t) ω − dξ 0 < x ≤ α ξ x α≤x ≤β v (x, t) = ω Z x ω(ξ, t) ω − dξ β ≤ x < 1. β 1−ξ Integrating by parts with respect to ξ , we obtain Z α v(ξ, t)dξ 0
which implies that Z α Z v(ξ, t)dξ = x
1 β
v(ξ, t)dξ = 0.
(28)
Then, from equality (27) we obtain Z Z ∂u N vdxdt = − A (t) uvdxdt, Ω Ω ∂t
(29)
where N v = φ(x)v,
and
A (t) u = −
∂ ∂x
φ(x)a (t)
∂u ∂x
.
If we introduce the smoothing operators with respect to t [18,22–24], Jξ−1 = I + ξ ∂t∂ operators provide the solutions of the respective problems: dgξ (t) + gξ (t) = g (t) dt gξ (t) |t=0 = 0
ξ
−1
∗ and Jξ−1 , then these
(30)
and dgξ∗ (t)
+ gξ∗ (t) = g (t) dt gξ (t) |t=T = 0
−ξ
(31)
∗ and also have the following properties: for any g ∈ L 2 (0, T ), the functions gξ = Jξ−1 g and gξ∗ = Jξ−1 g are 2 RT in W21 (0, T ) such that gξ |t=0 = 0 and gξ∗ |t=T = 0. Moreover, Jξ−1 commutes with ∂t∂ , so 0 gξ − g dt −→ 0 and 2 R T ∗ − g g dt −→ 0 for ξ −→ 0. ξ 0 Rt ξ a2 Putting u = 0 exp(cτ )vξ∗ (x, τ ) dτ in (29), where the constant c satisfies ca0 − a3 − 2a30 ≥ 0, and using (31), we obtain Z Z Z ∂vξ∗ ∂u dxdt. (32) − exp(ct)vξ∗ N vdxdt = + A (t) u exp(−ct) dxdt − ξ A (t) u ∂t ∂t Ω Ω Ω
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Integrating by parts each term in the left-hand side of (32) and taking the real parts yields 2 Z Z 1 ∂u ∂u A (t) u exp(−ct) dxdt = a (t) exp(−ct)φ(x) (x, T ) dx 2Re ∂t ∂t Ω 0 Z da(t) ∂u 2 exp(−ct)φ(x) ca (t) − + ∂ x dxdt, dt Ω ! ! Z Z ∗ ∂vξ∗ da(t) ∂u ∂vξ A (t) u Re −ξ dxdt = Re ξ φ(x) dxdt ∂t ∂x ∂x Ω Ω dt Z ∂v ∗ 2 ξ a (t) exp(−ct)φ(x) + dxdt. ∂x Ω
(33)
(34)
Using ε-inequalities we obtain Z Re −ξ Ω
A (t) u
∂vξ∗ ∂t
! dxdt
≥
Z da(t) 2 ∂u 2 −ξ dxdt. exp(−ct)φ(x) 2a0 Ω dt ∂ x
(35)
Combining (33) and (35) we get Z −Re Ω
exp(ct)vξ∗ N vdxdt
ξ a2 ≥ exp(−ct)φ(x) ca0 − a3 − 3 2a0 Ω Z
! ∂u 2 dxdt ≥ 0. ∂x
(36)
Now, using (36), we have Z Re exp(ct)vξ∗ N vdxdt ≤ 0. Ω
Then, for ξ −→ 0 we obtain Z Z Re exp(ct)v N vdxdt = exp(ct)φ(x) |v|2 dxdt ≤ 0. Ω
Ω
We conclude that v = 0; hence, ω = 0, which ends the proof of the lemma.
Theorem 6. The range R(L) of L coincides with F. Proof. Since F is a Hilbert space, we have R(L) = F if and only if the relation Z 1 Z 1 Z α Z θ (x) θ (x) dlu dϕ Lu f dxdt + dx + luϕdx + luϕdx = 0 2 2 dx dx Ω 0 0 β
(37)
for arbitrary u ∈ E and ( f, ϕ, ) ∈ F, implies that f = 0 and ϕ = 0. Putting u ∈ D0 (L) in (37), we conclude from the Lemma 3 that ψ f = 0, then f = 0, where 0
θ(x) dϕ 2 2 | dx |
+
Rα 0
|ϕ|2 dx
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